📚 Mastering Number and Algebra for IGCSE Edexcel Mathematics | 攻克 IGCSE Edexcel 数学：数字与代数
Number and algebra form the backbone of the IGCSE Edexcel Mathematics specification (4MA1). This article revisits the essential concepts you must master, from standard form and percentages to solving quadratic equations and interpreting inequalities. Each section provides clear steps, typical exam pitfalls, and practice-focused explanations to sharpen your problem-solving skills.
数字与代数是 Edexcel IGCSE 数学 (4MA1) 的基础核心。本文带你重温必须掌握的关键概念，涵盖标准形式、百分数、二次方程求解与不等式等重要内容。每个小节提供清晰步骤、常见易错点与讲练结合的解析，帮助你提升解题能力。
1. Understanding Standard Form | 理解标准形式
7 × 10⁻³ equals 0.0047. This compact notation helps avoid errors when handling many zeros.
标准形式将极大或极小的数字表示为 a × 10ⁿ，其中 1 ≤ a < 10，n 为整数。例如 3.2 × 10⁵ 代表 320 000，4.7 × 10⁻³ 等于 0.0047。这种紧凑表示法能有效避免处理多个零时的错误。
When adding or subtracting in standard form, convert both numbers to the same power of 10 first. Edexcel exam questions often combine standard form with unit conversions, so always check the exponent after multiplying or dividing.
进行加减运算时，先将两个数转换为相同幂次。Edexcel 考试题常将标准形式与单位换算结合，因此在乘除运算后务必核对指数是否正确。
2. Fractions, Decimals and Percentages | 分数、小数与百分数
tween fractions, decimals and percentages is a recurring skill. Remember that ½ = 0.5 = 50%, ⅓ ≈ 0.333... = 33⅓%. To compare quantities, convert them all to the same form—decimals often make ordering easiest.
分数、小数和百分数之间的相互转换是高频考点。记住 ½ = 0.5 = 50%，⅓ ≈ 0.333... = 33⅓%。比较数量大小时，可将它们全部转换为同一形式——通常换成小数排序最方便。
For percentage increase or decrease, use the multiplier method: a 15% increase multiplies by 1.15, and a 20% decrease multiplies by 0.80. Avoid the common mistake of applying percentage changes additively in repeated contexts.
处理百分比增减时，使用乘数法：增加 15% 即乘以 1.15，减少 20% 则乘以 0.80。避免在多次增减时将百分比直接相加减的常见错误。
3. Solving Linear Equations | 解线性方程
ax + b = c. Solve by performing inverse operations to isolate x. For example, 3x − 7 = 11 → add 7: 3x = 18 → divide by 3: x = 6. Always substitute your answer back to check.
线性方程具有 ax + b = c 的形式，通过逆运算分离变量 x 来求解。例如 3x − 7 = 11 → 加上 7：3x = 18 → 除以 3：x = 6。务必代入验证。
f fractions appear, multiply every term by the least common denominator to eliminate them.
含括号的方程先展开：2(x + 5) = 3(x − 1) → 2x + 10 = 3x − 3 → 解得 x = 13。若出现分数，将每一项乘以最小公分母消除分母。
4. Factorising and Solving Quadratic Equations | 因式分解与解二次方程
The goal is to write it as (px + q)(rx + s) = 0, then set each bracket to zero to find x.
形如 ax² + bx + c = 0 的二次方程常可通过因式分解求解。目标是将它写成 (px + q)(rx + s) = 0，然后令每个括号为零求 x。
se the ac method or trial and error. Remember that not all quadratics factorise nicely; in such cases use the quadratic formula.
对于 x² + 5x + 6 = 0，找出两数乘积为 6、和为 5：2 与 3，因此 (x + 2)(x + 3) = 0 → x = −2 或 x = −3。当 a ≠ 1 时，可使用 ac 法或试凑法。注意不是所有二次式都能顺利分解，此时应选用求根公式。
The quadratic formula is x = [−b ± √b² − 4ac] / 2a. Always evaluate the discriminant b² − 4ac first to determine the number of solutions. IGCSE questions may ask for answers correct to 3 significant figures.
求根公式为 x = [−b ± √b² − 4ac] / 2a。务必先计算判别式 b² − 4ac，以判断解的个数。IGCSE 试题可能要求答案精确至三位有效数字。
5. Working with Inequalities | 处理不等式
Inequalities behave like equations except when multiplying or dividing by a negative number, the inequality sign reverses. Solve 5 − 2x > 1: subtract 5 gives −2x > −4; divide by −2 gives x < 2.
不等式与方程运算相似，但当乘以或除以负数时，不等号方向反转。求解 5 − 2x > 1：减 5 得 −2x > −4；除以 −2 得 x < 2。
2x < 4, then divide by 2 → −2 ≤ x < 2. Number-line representations are frequently tested.
形如 −3 ≤ 2x + 1 < 5 的双重不等式需同时处理三部分：减 1 → −4 ≤ 2x < 4，再除以 2 → −2 ≤ x < 2。数轴表示是常见考点。
6. Sequence Patterns and the nth Term | 数列模式与第 n 项
uence 7, 11, 15, 19,... the difference is 4, so the formula is 4n + 3 (since 4×1 + 3 = 7).
线性数列具有恒定差值。第 n 项公式为 an + b，其中 a 为公差。数列 7, 11, 15, 19,... 的公差为 4，因此通项公式为 4n + 3（因为 4×1 + 3 = 7）。
For quadratic sequences, the second difference is constant. The nth term is of the form an² + bn + c. To find a, half the second difference; then use known terms to determine b and c. Practice identifying patterns in diagrams is essential.
二次数列的二阶差恒定。第 n 项形式为 an² + bn + c。将二阶差除以 2 得到 a，再代入已知项求 b 和 c。务必练习图形题中的规律识别。
7. Straight Line Graphs and Functions | 直线图与函数
The equation of a straight line is y = mx + c, where m is the gradient and c is the y‑ To find the gradient from two points (x₁, y₁) and (x₂, y₂), use m = (y₂ − y₁) / (x₂ − x₁).
直线方程为 y = mx + c，其中 m 为斜率，c 为 y 轴截距。由两点 (x₁, y₁) 和 (x₂, y₂) 求斜率，使用 m = (y₂ − y₁) / (x₂ − x₁)。
) = 2x + 3; composite and inverse functions appear in IGCSE, so practise fg(x) and f⁻¹(x).
平行线斜率相等；垂直线满足 m₁ × m₂ = −1。画图需明确截距与斜率。函数可表示为 f(x) = 2x + 3；IGCSE 会考查复合函数与反函数，务必练习 fg(x) 和 f⁻¹(x)。
8. Ratio, Proportion and Scales | 比、比例与尺度
viding all parts by their highest common factor. A ratio 24:36 simplifies to 2:3. To share £50 in the ratio 2:3, find the total parts (5) and work out 2/5 × £50 = £20 and 3/5 × £50 = £30.
化简比时将各项除以最大公约数。比 24:36 简化为 2:3。按 2:3 分配 £50，总份数为 5，计算 2/5 × £50 = £20，3/5 × £50 = £30。
nships from tables or descriptions is a core skill. Map scales and model scales also feature prominently; a scale of 1:50 000 means 1 cm on the map represents 50 000 cm (0.5 km) in reality.
正比例意味着 y = kx，k 为常数。反比例则为 y = k/x。通过表格或文字描述识别这类关系是核心技能。地图比例尺与模型比例也经常出现；比例尺 1:50 000 表示地图上 1 cm 代表实际 50 000 cm（0.5 km）。
9. Indices, Roots and Surds | 指数、根与根式
Negative indices negative reciprocals: a⁻ⁿ = 1/aⁿ. Fractional indices link to roots: a½ = √a, a⅓ = ∛a.
指数运算法则必不可少：aᵐ × aⁿ = aᵐ⁺ⁿ，aᵐ ÷ aⁿ = aᵐ⁻ⁿ，(aᵐ)ⁿ = aᵐⁿ。负指数表示倒数：a⁻ⁿ = 1/aⁿ。分数指数与根号相关：a½ = √a，a⅓ = ∛a。
A surd is an irrational root like √2. Simplify surds by identifying square factors: √72 = √36×2 = 6√2. Rationalising denominators such as 1/(√3 + 1) involves multiplying numerator and denominator by the conjugate.
根式是指像 √2 这样的无理根。通过提取平方因子化简：√72 = √36×2 = 6√2。分母有理化例如 1/(√3 + 1)，可利用共轭式进行分子分母同乘。
10. Word Problems and Algebraic Modelling | 应用题与代数建模
Many IGCSE questions embed algebra in real‑life contexts: ages, coins, area and perimeter, or distance‑speed‑time. Translate the words into algebraic expressions first, then form equations.
许多 IGCSE 题目将代数融入现实情境：年龄、硬币、面积周长或路程‑速度‑时间问题。先将其文字转化为代数式，再建立方程。
Example: 'Three years ago, Amy was twice as old as Ben. Their current ages sum to 36.' Let Amy's age = a, Ben' = 2(b − 3) and a + b = 36. Solve simultaneously to find ages. Diagrams help clarify shape‑based problems.
例如：“三年前 Amy 的年龄是 Ben 的两倍。两人现在的年龄和为 36。”设 Amy 年龄为 a，Ben 为 b，则 a − 3 = 2(b − 3) 且 a + b = 36。联立求解。绘制示意图有助于理清图形类问题。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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